NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

Updated on Jul 21, 2020 - 5:27 p.m. IST by Safeer PP

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement: As far as CBSE board is considered a good percentage of questions are asked directly from NCERT concepts and exercise. This is where a tool like CBSE NCERT solutions for class 11 physics chapter 2 units and measurement is important. NCERT solutions for class 11 gives an explanation to all the exercise questions. Measurements are always specified with a unit. Without units, measurement is meaningless. For example, a man tells the distance between my home and the nearest city is 5. This statement by the man is meaningless. Whether it is 5m, 5cm or 5Km makes it meaningful. The solutions of NCERT class 11 physics chapter 2 units and measurement start with questions related to the conversion of units. The concept of accuracy, precision and error can be studied well with the help of CBSE NCERT solutions for class 11 physics chapter 2 units and measurement. NCERT solutions help in self-evaluation of the concept studied in the chapter. The important topics of chapter 2 units and measurement are the concepts of errors and measurement of errors, dimensional analysis and significant figures. Questions from these topics are explained in NCERT solutions for class 11 physics chapter 2 units and measurement.

NCERT solutions for class 11 physics chapter 2 units and measurement exercises

Q2.1 Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ..... $m^3$

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... $\small (mm)^2$

(d) The relative density of lead is 11.3. Its density is .... $\small g cm^{-3}$ or .... $\small kg m^{-3}$ .

Answer:

(a) We know, $\small 1 cm = .01 m$ (Tip: Divide by 100 to convert cm to m)

The volume of a cube of side a = $\small a^3\ m^3$

Volume of cube of side 1 cm (i.e, .01 m ) = $\small (.01)^3 = 10^{-6}\ m^3$

(b) We know, $1 cm = 10 mm$ (Tip: Multiply by 10 to convert cm to mm)

The surface area of a solid cylinder of radius r and height h = $2(\pi r^2) + 2 \pi rh = 2\pi r(r + h)$

Required area = $2\pi (20)(\ 20 + 100) mm^3 = 40 \pi (120) mm^3 = 4800 \pi\ mm^3 = 1.5 \times 10^{4 }\ mm^3$

Ditance covered = $Speed \times time = (18 \times 5/18)ms^{-1} \times (1s) = 5 m$

(d) Density = Relative Density $\times$ Density of water

(Density of water = $1 \ gcm^{-3} = 1000\ kgm^{-3}$ )

Density of lead=

$(11.3)\times1\ gcm^{-3} = 11.3\ gcm^{-3} \\ (11.3)\times1000\ kgm^{-3} = 11300\ kgm^{-3} = 1.13 \times 10^4\ kgm^{-3}$

Q 2.2 : Fill in the blanks by suitable conversion of units

(a) $1 kg m^2s^-^2 = \ \ \ \ \ \ \ \ gcm^2s^-^2$

b) $1 m = \ \ \ \ ly$

(d) $G = 6.67 \times 10^{-11} Nm^2(kg)^-^2 = \ \ \ \ \ (cm)^3 s^{-2}g^{-1}$ .

Answer:

(a) $(1 kg \rightarrow 1000g ;\ 1m\rightarrow 100 cm)$

$1 kgm^2s^{-2} = (1000g)(100cm)^2s^{-2} = (10^3 \times10^{4})gcm^2s^{-2} = 10^{7}\ gcm^2s^{-2}$

(b) $1 m = 1.057\times10^{-16} ly$ (1 ly = Distance travelled by light in 1 year )

$\\3.0 ms^{-2} = 3.0 (10^{-3}km)(1/3600 hr)^{-2} = 3.0 \times (3600)^2 / 1000\ kmhr^{-2} \\ = 3.9 \times 10^4 \ kmhr^{-2}$

(d) $(N \rightarrow J.s = kgms^{-2};\ 1 m \rightarrow 100cm;\ 1 kg\rightarrow 1000g)$

$G = 6.67 \times 10^{-11} (kgms^{-2})m^2kg^-^2 = (6.67 \times 10^{-11})(m)^3 s^{-2}kg^{-1} = 6.67 \times 10^{-8}cm^3 s^{-2}g^{-1}$

Q 2.3 : A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where $1J = 1 kg m^2 s^{-2}$ . Suppose we employ a system of units in which the unit of mass equals $\alpha \ kg$ , the unit of length equals $\beta \ m$ , the unit of time is $\gamma \ s$ . Show that a calorie has a magnitude 4.2 $\alpha^{-1}\beta ^{-2}\gamma^{2}$ in terms of the new units.

Answer:

Given,

$1 Cal = 4.2 (kg)(m)^{2}(s)^{-2}$

Given new unit of mass = $\alpha kg$ (In old unit 1kg corresponded to a unit mass, but in new unit $\alpha kg$ corresponds to a unit mass)

$\therefore$ In terms of the new unit, $1 kg = 1/\alpha = \alpha ^{-1}$

Similarly in terms of new units , $1 m = 1/\beta = \beta^{-1}$ and $1 s = 1/\gamma = \gamma ^{-1}$

$\therefore$ $1 Cal = 4.2 (kg)(m)^{2}(s)^{-2} = 4.2 (\alpha ^{-1})(\beta ^{-1})^{2}(\gamma ^{-1})^{-2} = 4.2 \alpha ^{-1}\beta ^{-2}\gamma ^{2}$

Q 2.4 : Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :

(a) atoms are very small objects

(b) a jet plane moves with great speed

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light

Answer:

The given statement is true. A dimensional quantity may be small with respect to one reference and maybe large with respect to another reference. Hence, we require a standard reference to judge for comparison.

(a) An atom is a very small object with respect to a tennis ball. (but larger than an electron!)

(b) A jet plane moves with great speed with respect to a train.

(d) The air inside this room contains a large number of molecules as compared to in your lungs.

(e) A proton is much more massive than an electron

(f) The speed of sound is less than the speed of light

Q 2.5 : A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer:

Distance between Sun and Earth = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken by light to reach earth is 8 minute 20 seconds

Time taken = $(8\times60) + 20 = 500 s$

The distance between Sun and Earth = 1 x 500 = 500 units.

Q 2.6 : Which of the following is the most precise device for measuring length :

(a) a vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

Answer:

To judge which tool is more precise, we have to find out their least count. Least count defines the margin of error and hence the precision. Hence the instrument with lower least count will be more precise.

(a) Least count = $1MSD - 1VSD = 1 - (19/20)= 1/20 = 0.05\ cm$ (Taking 1 MSD as 1 mm)

(b) Least count = pitch/ number of divisions

= $1/10000 = 0.001 cm$

Therefore, the optical instrument is the most precise device used to measure length.

Q 2.7 : A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

Answer:

Given,

Magnification of Microscope = 100

The average width of hair under the microscope = 3.5 mm

(20 observations were made to calculate the average i.e. 3.5 mm as an experimental procedure. No need in our calculations.)

(Note: When magnified, the width is 3.5 mm. Hence actual width will be less by a factor of magnification value)

The average thickness of hair = $\frac{3.5 mm}{100} = 0.035 \ mm.$

Q2.8 Answer the following :

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Answer:

(a) Take the thread and wrap it around the metre scale. Make sure the coils are packed closely without any space in between. If the diameter of the thread is d and number of turns obtained are n, then (n x d) corresponds to the marking on the metre scale, l.

Therefore, the diameter of the thread would be, d = l/n

(b) Theoretically, by increasing the number of divisions on the circular divisions, the value of least count decreases and hence accuracy increases.(Lower the value of least count, better will be the reading)

But practically, the number of divisions can be increased only up to a certain limit. (Also two adjacent divisions cannot be separated by a distance less than the human eye resolution!)

(c) With an increase in the number of observations, the accuracy of the experiment increases as the error is now distributed over a large range. Hence, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements.

Q 2.9 : The photograph of a house occupies an area of 1.75 $cm^2$ on a 35 $mm$ slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 $m^2$ . What is the linear magnification of the projector-screen arrangement?

Answer:

Given,

Area of the house in the photo = $1.75\ cm^{2}$

Area of the house on the screen = $1.55 m^{2} = 1.55 \times10^{4} cm^2$

$\therefore$ Arial magnification, $m_{a}$ = Area on the screen / area on photo = $\frac{1.55\times 10^{4}}{1.75} = 0.886 \times 10^4$

Linear magnification of the projector- screen arrangement $m_{l} = \sqrt{m_{a}} = \sqrt{.886 \times 10^{4}}= 94.1$

Q 2.10 : State the number of significant figures in the following :
(a) $0.007 \ m^2$

(b) $2.64 \times 10^{24} kg$

(d) $6.320 \ J$

(e) $6.032 \ Nm^{-2}$

(f) $0.0006032 \ m^{2}$

Answer:

(a) The given value is $0.007 \ m^2$ .

Since, the number is less than 1 , the zeros on the right to the decimal before the first non-zero integer is insignificant. So, the number 7 is the only significant digit.

$\therefore$ It has 1 significant digit.

(b) The value is $2.64 \times 10^{24} kg$

For the determination of significant values, we do not consider the power of 10 (Number is not less than 1). The digits 2, 6, and 4 are significant figures.

$\therefore$ It has 3 significant digits.

(c)The value is $0.2370 \ g cm^{-3}$ .

For the given value with decimals, all the numbers 2, 3, 7 and 0 are significant.

$\therefore$ It has 4 significant digits.

(d) The value is $6.320 \ J$ .

It has 4 significant digits.

(e) The value is $6.032 \ Nm^{-2}$ .

All the four digits are significant as the zeros in between two non-zero values are also significant.

$\therefore$ It has 4 significant digits.

(f) The value is $0.0006032 \ m^{2}$

Same as (a), first three zeroes after the decimal is insignificant. Only 6, 0, 3, 2 are significant.

$\therefore$ It has 4 significant digits.

Q 2.11 : The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer:

Given,

Length, l = 4.234 m ; Breadth, b = 1.005 m; Height, h = 2.01 cm = 0.0201 m

The length has 4 significant figures

The breadth has 4 significant figures

The height has 3 significant figures (Since the number is less than 1, hence zeroes after decimal before the first non-zero integer is insignificant)

We know,

Surface area of a cuboid = 2(l x b + b x h + h x l)

= $2(4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234)$

= $2(4.255 + 0.020 + 0.085)$

= $8.72 \ m^{2}$ (Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

Volume = l x b x h

= $4.234 \times 1.005 \times 0.0201$ = $0.0855 \ m^3$ ( $\therefore$ 3 significant digits)

(Note: For Multiplication The LEAST number of significant digits in any number determines the number of significant figures in the answer; i.e decided by the number of significant digits )

The area has three significant values 2, 7 and 8.

The volume has three significant values 5, 5 and 8.

Q 2.12 : The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Answer:

Given,

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g = 0.02015 kg

The mass of the second gold piece = 20.17 g = 0.02017 kg

(a) The total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg

Since one is the least number of decimal places, the total mass = 2.3 kg.

(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

Q 2.13 : A physical quantity P is related to four observables a, b, c and d as follows : $P = a^3b^2/(\sqrt{c}d)$ The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Answer:

Given, $P = a^3b^2/(\sqrt{c}d)$

(a) $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}$

$\frac{\Delta P}{P} \times 100 \%= (3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}) \times 100 \%$

$\ \ \ \ \ \ \ = (3\times1 + 2\times3 + .5\times4 + 1\times2)\% = 13\%$ $(\Delta a/a = 1\% = 1/100)$

$\therefore$ The percentage error in the quantity P = 13 %

(b) Rounding off the value of P = 3.8

2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :

(a) $y = a sin 2\pi t/T$

(b) $y = a sin vt$

(d) $y = ( a\sqrt{2}) (sin 2\pi t / T+ cos 2\pi t / T )$

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Answer:

Ground rules:

$\\sin\Theta , cos\Theta \ are\ DIMENSIONLESS.\ \\ and\ \Theta\ must\ be\ dimensionless$

[y] = L ( $M^0 L^1 T^0$ )

[a] = L

[v] = $LT^{-1}$

[t/T] is Dimenionless.

(a) The dimensions on both sides are equal, the formula is dimensionally correct .

(b) $\because$ [vt] = ( $LT^{-1}$ )(T) = L ( $\therefore \Theta$ is not dimensionless)

The formula is dimensionally incorrect

It is dimensionally incorrect , as the dimensions on both sides are not equal.

(d) The dimensions on both sides are equal, the formula is dimensionally correct . (Don't get confused by summation of trigonometric functions !)

Q 2.15 : A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_{o}$ of a particle in terms of its speed $v$ and the speed of light, $c$ . (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Guess where to put the missing c.

Answer:

The relation given is $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Divide both sides by $m_{o}$ ; $\therefore$ L.H.S becomes $m / m_{o}$ which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by $m_{o}$ !)

$\frac{1}{(1-v^2)^{1/2}}$ can be dimensionless only when $v \rightarrow (v/c)$

Therefore, the dimensional equation is $m = \frac{m_{o}}{(1-(\frac{v}{c})^2)^{1/2}}$

Q 2.16 : The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\AA : 1\ \AA = 10^{-10} m$ . The size of a hydrogen atom is about $0.5 \AA$ . What is the total atomic volume in $m^3$ of a mole of hydrogen atoms?

Answer:

Radius of an Hydrogen atom = 0.5 $\AA$ = 0.5 x 10 ^-10 m

Volume = $\frac{4}{3} \pi r^3$

= $4/3 \times 22/7 \times (0.5 \times 10^{-10})^3$

= $0.524 \times 10^{-30} m^3$

1 hydrogen mole contains $6.023 \times 10^{23}$ hydrogen atoms.

The volume of 1 mole of hydrogen atom = $6.023 \times 10^{23} \times 0.524 \times 10^{-30}$

= $3.16 \times 10^{-7} m^3 \approx 3 \times 10^{-7} m^3$ .

Q 2.17 : One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer:

Radius of hydrogen atom = 0.5 $\AA$ = 0.5 x 10 ^-10 m (Size here refers to Diameter!)

Volume occupied by the hydrogen atom= $\frac{4}{3} \pi r^3$

= $4/3 \times 22/7 \times ( 0.5 \times 10^{-10})^3$

= $0.524 \times 10^{-30} m^3$

1 mole of hydrogen contains 6.023 x 10 ²³ hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10 ²³ x 0.524 x 10 ^-30

= 3.16 x 10 ^-7 m ³

$V_{m} = 22.4 L = 22.4 \times 10^{-3} m^3$

$\frac{V_{m}}{V_{a}} = \frac{22.4\times10^{-3}}{3.16\times10^{-7}} = 7.09 \times 10^4$

The molar volume is $7.09 \times 10^4$ times greater than the atomic volume.

Hence, intermolecular separation in gas is much larger than the size of a molecule.

2.18 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer:

Our eyes detect angular velocity, not absolute velocity. An object far away makes a lesser angle than an object which is close. That's why the moon (which is so far away!) does not seem to move at all angularly and thus seems to follow you while driving.

In other words, while in a moving train, or for that matter in any moving vehicle, a nearby object moves in the opposite direction while the distant object moves in the same direction. !

2.19 The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ $3 \times 10^{11} m$ . However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of $1{}''$ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of $1{}''$ (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer:

The diameter of Earth’s orbit = $3 \times 10^{11} m$
$\therefore$ The radius of Earth’s orbit, r = $1.5 \times 10^{11} m$
Let the distance parallax angle be = $4.847 \times 10^{-6} rad$ .
Let the distance between earth and star be R.
(Parsec is the distance at which average radius of earth’s orbit
subtends an angle of $1{}''$ .)
We have $\Theta = r/R$ (Analogous to a circle, R here is the radius, r is the arc length and $\Theta$ is the angle covered ! )

$R = \frac{r}{\Theta } = \frac{1.5 \times 10^{11}}{4.847 \times 10^{-6}} = 0.309 \times 10^{17}$

$= 3.09 \times 10^{16}\ m$

Hence, 1 parsec $= 3.09 \times 10^{16}\ m$ .

Q 2.20 : The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answer:

Given, Distance of the star from the solar system = 4.29 ly (light years)
1 light year is the distance travelled by light in one year.

(Note: Light year is a measurement of distance and not time!)
(a) $1 ly = (3 \times 10^8)ms^{-1} \times (365 \times 24 \times 60 \times 60)s = 94608 \times 10^{11} m$
4.29 ly = $405868.32 \times 10^{11} m$
We know, 1 parsec = $3.08 \times 10^{16} m$

$\therefore$ 4.29 ly = $\frac{405868.32\times10^{11}}{3.08\times10^{16}}$ = 1.32 parsec

Now,

(b) $? = d/D$ $\Theta = d/D$

& d = $3 \times 10^{11} m$ ; D = $405868.32 \times 10^{11} m$

$\theta = (3\times10^{11})/(405868.32\times10^{11} )= 7.39\times10^{-6} rad$

Also, We know $1\ sec = 4.85 \times10^{-6} rad$

$\therefore 7.39\times10^{-6} rad = \frac{7.39\times10^{-6}}{4.85 \times10^{-6}} = 1.52''$

Q 2.21 : Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer:

The statement "Precise measurements of physical quantities are a need of science" is indeed true. In Space explorations, very precise measurement of time in microsecond range is needed. In determining the half-life of radioactive material, a very precise value of the mass of nuclear particles is required. Similarly, in Spectroscopy precise value of the length in Angstroms is required.

Q 2.22 : Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :

(a) the total mass of rain-bearing clouds over India during the Monsoon

(b) the mass of an elephant

(d) the number of strands of hair on your head

(e) the number of air molecules in your classroom.

Answer:

(a) Height of water column during monsoon is recorded as 215 cm.

H = 215 cm = 2.15 m

Area of the country, $A = 3.3 \times 10^{12} m^2$

Volume of water column, V = AH

V = $3.3 \times 10^{12} m^2 \times 2.15 m = 7.1 \times 10^{12} m^3$

Mass of the rain-bearing clouds over India during the Monsoon, m = Volume x Density

m = $7.1 \times 10^{12} m^3 \times 10^3 kg m^{-3}$ = ⁽ Density of water = 10 ³ kg m ^{-3 )}

b) Consider a large solid cube of known density having a density less than water.

Measure the volume of water displaced when it immersed in water = v

Measure the volume again when the elephant is kept on the cube = V

The volume of water displaced by elephant, V' = V – v

The mass of this volume of water is equal to the mass of the elephant.

Mass of water displaced by elephant, m = V' x Density of water

This gives the approximate mass of the elephant.

(c) A rotating device can be used to determine the speed of the wind. As the wind blows, the number of rotations per second will give the wind speed.

(d) Let A be the area of the head covered with hair.

If r is the radius of the root of the hair, the area of the hair strand, $a = \pi r^2$

So, the number of hair , $n = A/a = A/\pi r^2$

(e) Let l, b and h be the length, breadth and height of the classroom, $\therefore$ Volume of the room, v = lbh.

The volume of the air molecule, $v' = (4/3)\pi r^3$ (r is the radius of an air molecule)

So, the number of air molecules in the classroom, $n = v'/v = 4\pi r^3/3lbh$

Q 2.23: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7 K$ , and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = $2.0 \times 10^{30} kg$ , radius of the Sun = $7.0 \times 10^8 m$

Answer:

Given,

Mass of the Sun, m = $3 \times10^{30}\ kg$

The radius of the Sun, r = $8 \times10^8\ m$

$\therefore$ Volume V = $\frac{4}{3}\pi r^3$

= $\frac{4}{3} \times \frac{22}{7} \times (8 \times 10^8)^3 = 2145.52 \times10^{24} m^3$

Density = Mass/Volume

= $\frac{3\times10^{30}}{2145.52\times10^{24}} = 1.39 \times 10^3\ kgm^{-3}$

Therefore, the density of the sun is in the range of solids and liquids and not gases. This high density arises due to inward gravitational attraction on outer layers due to inner layers of the Sun. (Imagine layers and layers of gases stacking up like a pile!)

Q 2.24: When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

Answer:

Given,

The distance of Jupiter, D = $824.7 \times10^6\ km$

Angular diameter, $\Theta = 35.72''$ $= 35.72 \times 4.848 \times 10^{-6} rad$ ( $\because 1'' = 4.848 \times 10^{-6} rad$ )

Let diameter of Jupiter = d km

$\\ \therefore d = \theta \times D = 824.7 \times 10^6 \times 35.72 \times 4.848 \times 10^{-6} \\ = 1.428 \times 10^{5}\ km$ $(\because \theta = \frac{d}{D})$

NCERT solutions for class 11 physics chapter 2 units and measurement additional exercises

Q 2.25 : A man walking briskly in rain with speed v must slant his umbrella forward making an angle $\Theta$ with the vertical. A student derives the following relation between $\Theta$ and $v: tan\Theta = v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \Theta \rightarrow 0$ , as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.

Answer:

The derived formula $tan\Theta = v$ is dimensionally incorrect.

We know, Trigonometric functions are dimensionless.

Hence , [ $tan\Theta$ ] = $M^0L^0T^0$

and [v] = $M^0L^1T^{-1} = LT^{-1}$ .

$\therefore$ To make it dimensionally correct, we can divide v by $v_{r}$ (where $v_{r}$ is the speed of rain)

Thus, L.H.S and R.H.S are both dimensionless and hence dimensionally satisfied.

The new formula is : $tan\Theta = v / v_{r}$

Q 2.26 : It is claimed that two caesium clocks if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard caesium clock in measuring a time-interval of 1 s?

Answer:

In terms of seconds, 100 years = $100 \times 365\times 24\times 60\times60 = 3.154 \times10^{9}\ s$

Given, Difference between the two clocks after 100 years = 0.02 s

$\therefore$ In 1 s, the time difference $= \frac{0.02}{3.15\times10^{9}} = 6.35 \times 10^{-12} s$

$\therefore$ Accuracy in measuring a time interval of 1 s =

$\frac{1}{6.35 \times 10^{-12}} = 1.57 \times 10^{11} \approx 10^{11}$

$\therefore$ Accuracy of 1 part in $10^{11}\ to\ 10^{12}$

Q 2.27 : Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : $970 \ kg\ m^{-3}$ . Are the two densities of the same order of magnitude ? If so, why ?

Answer:

Radius, r = $0.5 \times 2.5 ï¿½ = 1.25 \times 10^{-10}\ m$

Now, Volume occupied by each atom $V = (4/3)\pi r^3 = (4/3)\times (22/7)\times(1.25\times 10^{-10})^3$

$= 8.18 \times 10^{-30} m^3$

We know, One mole of sodium has $6.023 \times 10^{23}$ atoms and has a mass of $23 \times 10^{-3} kg$

$\therefore$ Mass of each Sodium atom $= \frac{23\times10^{-3}}{6.023\times10^{23}} kg$

$\therefore$ Density = Mass/ Volume $= (\frac{23\times10^{-3}}{6.023\times10^{23}}\ kg )/ (8.18 \times 10^{-30} m^3) = 466.8 \times 10^{1}\ kgm^{-3}$

But, the mass density of sodium in its crystalline phase = $970 \ kg\ m^{-3}$

$\therefore$ The densities are almost of the same order. In the solid phase, atoms are tightly packed and thus interatomic space is very small.

Q 2.28 : The unit of length convenient on the nuclear scale is a fermi : $1 f = 10^{-15} m$ . Nuclear sizes obey roughly the following empirical relation : $r = r_{o} A^{1/3}$ where r is the radius of the nucleus, A it's mass number, and $r_{o}$ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer:

The equation for the radius of the nucleus is given by,

$r = r_{0} A^{1/3}$

The volume of the nucleus using the above relation, $\small V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r_{0}A^{1/3})^3 = \frac{4}{3}\pi Ar_{0}^3$

We know,

Mass = Mass number× Mass of single Nucleus

= $\small A\times1.67\times10^{-27} kg$ (given)

$\small \therefore$ Nuclear mass Density = Mass of nucleus/ Volume of nucleus =

$\small \frac{A\times1.67\times10^{-27} kg}{\frac{4}{3}\pi Ar_{0}^3\ m^3} =\frac{3\times1.67\times10^{-27}}{4\pi r_{0}^3\ }\ kgm^{-3}$

The derived density formula contains only one variable, $\small r_{0}$ and is independent of mass number A. Since $\small r_{0}$ is constant, hence nuclear mass density is nearly constant for different nuclei.

∴ The density of the sodium atom nucleus = $\small 2.29\times 10^{17} kgm^{-3} \approx 0.3\times 10^{18} kgm^{-3}$

Comparing it with the average mass density of a sodium atom obtained in Q 2.27. (Density of the order $\small 10^3 kgm^{-3}$ )

Nuclear density is typically $\small 10^{15}$ times the atomic density of matter!

Q 2.29 : A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?

Answer:

2.56 s is the total time taken by the LASER to reach Moon and again back Earth.

$\therefore$ Time taken by LASER to reach Moon = $\frac{1}{2}\times 2.56 s = 1.28 s$

We know, Speed of light $\approx 3\times10^8 ms^{-1}$

$\therefore$ The radius of the lunar orbit around the Earth = distance between Earth and Moon =

= Speed of light x Time taken by laser one-way = $3\times10^8\ ms^{-1}\times1.28\ s = 3.84\times 10^8 \ m$

Q 2.30 : A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = $1450\ m s^{-1}$ ).

Answer:

Given,

77.0 s is the total time between the generation of a probe wave and the reception of its echo after reflection.

$\therefore$ Time taken by sound to reach the enemy submarine = Half of the total time = $\frac{1}{2}\times 77 s = 38.5 s$

The distance of enemy ship = Speed of sound x Time taken to reach the submarine

= $1450\ ms^{-1} \times 38.5\ s = 55825\ m = 55.8\ km$

Q 2.31 : The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Answer:

Let the distance of a quasar from Earth be D km.

We know , Speed of light = $3\times10^8 ms^{-1}$

And, time taken by light to reach us , t = 3.0 billion years = $3\times 10^{9} years = 3\times 10^{9} \times365\times24\times60\times60 \ s$

$\therefore$ D =Speed of Light x t

= $(3 \times 10^8 ms^{-1}) \times(3\times 10^{9} \times365\times24\times60\times60 \ s)$

= $2.8\times 10^{22}\ m$

Q 2.32 : It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer:

From Examples 2.3 and 2.4 , we have,

The diameter of the Earth = $1.276 \times 10^7 m$
Distance between the Moon and the Earth, $D_{moon}$ = $3.84\times10^8 m$

Distance between the Sun and the Earth, $D_{sun}$ = $1.496\times10^{11} m$

The diameter of the Sun, $d_{sun}$ = $1.39\times10^9 m$

Let, Diameter of the Moon be $d_{moon}$

Now, During Solar eclipse, the angle subtended by Sun's diameter on Earth = angle subtended by moon's diameter

$\frac{1.39\times10^9}{1.496\times10^{11}} = \frac{d_{moon}}{3.83\times10^8}$ $(\because \Theta = d/D)$

$\Rightarrow d_{moon} = \frac{1.39\times10^9}{1.496\times10^{11}}\times 3.83\times10^8 = 3.56\times10^6 m$

Therefore, the diameter of the moon = $3.56\times10^3 km$

Q 2.33 : A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of ). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Answer:

$T = \frac{e^4}{16\ \pi^2 \ \epsilon _{0}^{2}\ m_{p} m_{e}^2\ c^3\ G } \ s$

The above equation consisting of basic constants of atomic physics and the gravitational constant G has the dimension of time.

e = charge of Electron = $-1.6\times10^{-19} C$

$\epsilon _{0}$ = absolute permittivity = $8.85\times10^{-12} F/m$

$m_{p}$ = Mass of the Proton = $1.67\times10^{-27} kg$

$m_{e}$ = Mass of the Electron = $9.1\times10^{-31} kg$
c = Speed of light in vacuum = $3\times10^8 m/s$
G = Universal Gravitational constant = $6.67\times10^{11} Nm^2kg^{-2}$

Considering T as the age of the universe and putting the values of the constants, we get:

$\therefore T \approx 6\times10^9\ years$

The age of the universe ≈ 6 billion years.!

NCERT solutions for class 11 physics chapter wise

Chapter 1	NCERT solutions for class 11 physics chapter 1 Physical world
Chapter 2	NCERT solutions for class 11 physics chapter 2 Units and Measurement
Chapter 3	NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line
Chapter 4	NCERT solutions for class 11 physics chapter 4 Motion in a Plane
Chapter 5	NCERT solutions for class 11 physics chapter 5 Laws of Motion
Chapter 6	NCERT solutions for class 11 physics chapter 6 Work, Energy and Power
Chapter 7	NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion
Chapter 8	NCERT solutions for class 11 physics chapter 8 Gravitation
Chapter 9	NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids
Chapter 10	NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids
Chapter 11	NCERT solutions for class 11 physics chapter 11 Thermal Properties of Matter
Chapter 12	NCERT solutions for class 11 physics chapter 12 Thermodynamics
Chapter 13	NCERT solutions for class 11 physics chapter 13 Kinetic Theory
Chapter 14	NCERT solutions for class 11 physics chapter 14 Oscillations
Chapter 15	NCERT solutions for class 11 physics chapter 15 Waves

NCERT solutions for class 11 Subject wise

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Use of NCERT solutions for class 11 physics chapter 2 units and measurement is important:

This is an easy and important chapter in the NCERT for annual exam as well as competitive exam point of view.
For competitive exams like NEET, JEE Mains, JEE advance, NEET and KVPY etc the CBSE NCERT solutions for class 11 physics chapter 2 units and measurement is important.
One question can be expected from this chapter in NEET and JEE Mains too.
The solutions of NCERT class 11 physics chapter 2 units and measurement helps to understand the chapter well and also to score well in class 11 exams.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

Question: What is the important of the chapter in the class 11 NCERT syllabus

Answer:

Units and measurements is an important chapter for the entire class 11 and 12 physics course. In all the chapters the concepts of dimensions are used.

Question: What is the important of the class 11 chapter 2 for JEE Main

Answer:

For JEE Main one questions can be expected from the chapter units and measurements

Question: How many questions are asked from units and measurements for NEET

Answer:

One or 2 questions can be expected from class 11 NCERT chapter units and measurements

Question: What are the important topics in units and measurements

Answer:

Dimension and dimensional analysis, errors and significant figures are important topics from units and measurements

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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

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Use of NCERT solutions for class 11 physics chapter 2 units and measurement is important:

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

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